"However, the probability of winning by always switching is a logically distinct concept from the probability of winning by switching given that the player has picked door 1 and the host has opened door 3." [1]
[1] https://en.wikipedia.org/wiki/Monty_Hall_problem#Criticism_o...
People pretty much always understand what the options are, but still get the analysis wrong. That is what I find interesting. Not the idea that there's some secret mechanism at play that alters the odds of a specific case. Even in that wonky "open the door on the right when possible" world, a contestant that always switches will still win 2/3 of the time.
There is a confirmation bias that people expect people who are aware of the Monty Hall Problem expect others to be wrong about it and attribute it to people not understanding. I think it's perfectly acceptable for someone to read the vague "the host, who is well-aware of what’s going on behind the scenes" and not assume that means he chooses to pick a goat. They're probably right, and the people saying 50% are reasoning along the lines of "there are two doors so 50/50".
For rigor, I think the host's rules need to be more strict: "The host knows what is behind the doors and will never reveal the car".
Or a 50% answer is correct if you lay out the principles of how the host acts given his knowledge.
It doesn’t matter that the host meant to do that. Now that the information is revealed, the probabilities have changed.
Obviously if the host revealed the car, that would also affect the probabilities!
But we are talking about Monty Hall after all. So I could be wrong :P
P(initial choice is correct) = 1/3
P(host shows goat) = 2/3
P(host shows goat | initial choice is correct) = 1
and vice versa P(host shows goat | initial choice is wrong) = 1/2
Applying Bayes:
P(initial choice is correct | host shows goat) = P(host shows goat | initial choice is correct) * P(initial choice is correct) / P(host shows goat) = 1 * 1/3 / (2/3) = 1/2
So, initial choice now has 1/2 odds for being right. The host revealing a goat gave us information on whether he was playing the P=1/2 or the P=1 game.
Intuitively applying this with the 100 door version: the host getting lucky and revealing 98 goats in a row makes it fairly likely that the initial door was correct all along and they didn't get super lucky avoiding the car 98 times, as there's a good chance the contestant helped them by hiding the car.
If you are interested in more details, Wikipedia lists this as the "Ignorant Monty" variant.
It puzzles me that so many smart people got it wrong. It’s so easy to check your working.
When I first heard of this problem, I got it wrong too. Then I was told the answer, and to prove it to myself, it’s trivial to list the scenarios and simulate on paper. It’s still uncomfortable to think about sometimes, but I know how to show it’s true.
But all these people never thought to just do the working? Crazy!